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Where is the gradient of phi a) perpendicular b) parallel to the z-axis?

Mathematics
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I guess I have to set something zero for a) and perhaps in b) z should be constant?
use dot product. If need more help tell. :)

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Other answers:

I do the dot thing to get the parellel part and a cross product for the perpendicular part?
yes. to find where it is perpendicular to z axis solve this equation: nabla f . (0,0,1)=0 for paralel : nabla f . (0,0,1)= |nabla f|
other way would be finding maximum and minimum of |nabla f|
So for the parallel part I dont use a any vector but the vector (0,0,1) and dot this vector with nabla?
for a) I ended up with z^2=xy Is that my final answert for a set of points for which the gradient is parallel to the z-axis?
for b) I got now y^2-xz=x^2-yz This looks also not very good as a final answer.
for b): 3(x^2-yz,y^2-xz,z^2-xy).(0,0,1)=3sqrt((x^2-yz)^2+(y^2-xz)^2+(z^2-xy)^2)=> (z^2-xy)=sqrt((x^2-yz)^2+(y^2-xz)^2+(z^2-xy)^2) now should try to simplify this somehow
noticing that it will happen when : (x^2-yz)^2=0 and (y^2-xz)^2=0
so x^2=yz and y^2=xz from 1º: y=x^2/z putting this in 2º: x^4/z^2=xz => x^3=z^3 =>x=z=y which is a line passing through the origin forming 45º with every axis
So x, y, and z can be any number but as long as the have all the same value the gradient of phi is parallel to the z-axis?
:( c) When is the gradient zero. There I already calculated that the gradient would be zero for x=y=z so it seems kind of weird that this is the same solution for the gradient to be parallel to the z-axis.
But you calculated x=y=z so how can x=y=0 and z= anything?
Ya that's true. Forget it.
On the other hand as much as I remmeber vector that is 0 is also considered perpendicular or parallel to anything
This is confusing. Did I at least got a) right? I ended up with z^2=xy but that looks kind of weird for a final answert, but I don´t know if I can simplify or clarify that anymore.
Yes it's right. Here you got the graph of that: http://www.wolframalpha.com/input/?i=z%5E2%3Dxy+
I think part b) is ok too. Not sure tough, how to explain that it is same as for c)
Hmmmm ... I am still pretty confused but if you think this is alright than I will stick to it. Thank you for your help, especially since this took so long. Thanks a lot, myko.
Glad to help. Sry for not clarifying it till the end...

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